In: Statistics and Probability
How to Run descriptives and a two-tailed, two sample assuming equal variance t-test.
Here's your data:
Weight of Apples in Grams | |||
Apple ID | Farm A | Farm B | |
1 | 131 | 151 | |
2 | 147 | 159 | |
3 | 134 | 162 | |
4 | 134 | 158 | |
5 | 136 | 159 | |
6 | 137 | 160 | |
7 | 140 | 150 | |
8 | 134 | 160 | |
9 | 136 | 160 | |
10 | 133 | 160 | |
11 | 134 | 160 | |
12 | 132 | 158 | |
13 | 139 | 162 | |
14 | 136 | 160 | |
15 | 135 | 154 | |
16 | 135 | 155 | |
17 | 135 | 159 | |
18 | 138 | 151 | |
19 | 134 | 151 | |
20 | 149 | 150 | |
21 | 149 | 155 | |
22 | 135 | 150 | |
23 | 148 | 150 | |
24 | 135 | 158 | |
25 | 149 | 152 | |
26 | 137 | 162 | |
27 | 134 | 158 | |
28 | 140 | 161 | |
29 | 138 | 160 | |
30 | 133 | 161 | |
31 | 140 | 158 | |
32 | 138 | 160 | |
33 | 150 | 149 | |
34 | 135 | 160 | |
35 | 136 | 160 | |
36 | 148 | 157 | |
37 | 150 | 152 | |
38 | 136 | 150 | |
39 | 142 | 157 | |
40 | 132 | 149 | |
41 | 133 | 160 | |
42 | 147 | 150 | |
43 | 135 | 159 | |
44 | 140 | 153 | |
45 | 132 | 151 | |
46 | 132 | 160 | |
47 | 136 | 151 | |
48 | 141 | 162 | |
49 | 136 | 162 | |
50 | 135 | 160 | |
51 | 142 | 160 | |
52 | 135 | 150 | |
53 | 132 | 158 | |
54 | 140 | 160 | |
55 | 136 | 152 | |
56 | 138 | 160 | |
57 | 132 | 159 | |
58 | 136 | 150 | |
59 | 136 | 161 | |
60 | 135 | 158 | |
61 | 135 | 152 | |
62 | 136 | 154 | |
63 | 137 | 151 | |
64 | 138 | 150 | |
65 | 137 | 150 | |
66 | 136 | 153 | |
67 | 134 | 153 |
as the data is large enough i am using MINITAB to solve the problem.
steps:-
copy the data in minitab stat basic statistics 2 sample t choose each sample in its own column in sample 1 select 'farm A' and in sample 2 select 'farm B' click options type 95 in confidence interval type 0.0 in hypothesized difference choose the alternative hypothesis difference hypothesized difference tick assume equal variance ok ok.
your output be :-
Two-Sample T-Test and CI: Farm A, Farm B
Method
μ₁: mean of Farm A |
µ₂: mean of Farm B |
Difference: μ₁ - µ₂ |
Equal variances are assumed for this analysis.
Descriptive Statistics
Sample | N | Mean | StDev | SE Mean |
Farm A | 67 | 137.55 | 5.02 | 0.61 |
Farm B | 67 | 156.07 | 4.43 | 0.54 |
Estimation for Difference
Difference | Pooled StDev |
95% CI for Difference |
-18.522 | 4.735 | (-20.141, -16.904) |
Test
Null hypothesis | H₀: μ₁ - µ₂ = 0 |
Alternative hypothesis | H₁: μ₁ - µ₂ ≠ 0 |
T-Value | DF | P-Value |
-22.64 | 132 | 0.000 |
decision:-
p value = 0.000 <0.05
so, we reject the null hypothesis.
we conclude that,
there is sufficient evidence to claim that there is a difference between weight of apples of farm A and farm B at 0.05 level of significance.
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